\documentclass[UTF8]{ctexart}
\title{数学公式}
\author{fosonmeng}
\begin{document}
	\maketitle
	\tableofcontents
	
	\section{导数公式}
	\subsection{初等函数}
	\begin{enumerate}
		\item $\left(C\right)' = 0$
		\item $\left(x^{\mu}\right)' = {\mu}x^{\mu - 1}$
		\item $\left({\sin} x\right)' = {\cos} x$
		\item $\left(\cos x\right)' = - \sin x$
		\item $\left(\tan x\right)' = \sec^{2} x$
		\item $\left(\cot x\right)' = - \csc^{2} x$
		\item $\left(\sec x\right)' = \sec x\tan x$
		\item $\left(\csc x\right)' = -csc x\cot x$
		\item $\left(a^{x}\right)' = a^{x}\ln a$
		\item $\left(e^{x}\right)' = e^{x}$
		\item $\left(\log_{a}x\right)' = \frac{1}{x\ln a}$
		\item $\left(\ln x\right)' = \frac{1}{x}$
		\item $\left(\arcsin x\right)' = \frac{1}{\sqrt{1 - x^{2}}}$
		\item $\left(\arccos x\right)' = - \frac{1}{\sqrt{1 - x^{2}}}$
		\item $\left(\arctan x\right)' = \frac{1}{1 + x^{2}}$
		\item $\left(arccot x\right)' = - \frac{1}{1 + x^{2}}$
	\end{enumerate}

	\subsection{和差积商}
	\begin{enumerate}
		\item $\left(u \pm v\right)' = u' \pm v'$
		\item $\left(Cu\right)' = Cu'$ (C是常数)
		\item $\left(uv\right)' = u'v + uv'$
		\item $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}$ ($v \ne 0$)
	\end{enumerate}

	\subsection{反函数求导}
	设$x = f\left(y\right)$在区间$I_{y}$内单调、可导且$f'\left(y\right) \ne 0$，则它的反函数
	$y = f^{-1}\left(x\right)$在$I_{x} = f\left(I_{y}\right)$内也可导，且\\
	\begin{center}
		$\left[f^{-1}\left(x\right)\right]' = \frac{1}{f^{-1}\left(y\right)}$ 或
		$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$
	\end{center}

	\subsection{复合函数求导}
	设$y = f\left(u\right)$，而$u = g\left(x\right)$且$f\left(u\right)$及$g\left(x\right)$都可导，则复合函数$y = f\left[g\left(x\right)\right]$的导数为\\
	\begin{center}
		$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ 或\\
		$y'\left(x\right) = f'\left(u\right){\cdot}g'\left(x\right)$
	\end{center}
\end{document}